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achapt 14

# achapt 14 - LINEAR ALGEBRAIC EQUATIONS 14 LINEAR ALGEBRAIC...

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Unformatted text preview: LINEAR ALGEBRAIC EQUATIONS 14 LINEAR ALGEBRAIC EQUATIONS Systems of linear algebraic equations arise in all walks of life. They represent the most basic type of system of equations and they’re taught to everyone as far back as 8-th grade. Yet, the complete story about linear algebraic equations is usually not taught. What happens when there are more equations than unknowns or fewer equations than unknowns? What happens when some of the equations are repeated? These are precisely the questions that are answered in this chapter. Before proceeding with this, we have some background material to learn. We first need to discuss ways to minimize a function of several variables. Then, we’ll need to understand how to do this using a matrix-vector notation. After this is done, we’ll look at linear algebraic equations. 1. How to Minimize a Function of Several Variables The best way to introduce this topic is with an example. Let’s minimize the function MAE 461: DYNAMICS AND CONTROLS LINEAR ALGEBRAIC EQUATIONS (14 - 1) 2 2 y x J + = where x and y are both constrained to lie on the line (See Fig. 14 – 1) (14 - 2) b ax y + = Figure 14 - 1 This problem amounts to finding the point on the line that is closer to the origin than any other point. There are several ways to solve this problem. Let’s look at the following three ways. Ordinary Geometry The first way to solve this problem is to use ordinary geometry. Draw a perpendicular to the line that intersects the origin. The equation of the perpendicular line is y = - x / a (See Fig. 14 – 2). MAE 461: DYNAMICS AND CONTROLS LINEAR ALGEBRAIC EQUATIONS Figure 14 - 2 Substituting the equation for the perpendicular line into the equation for the line yields the intercepts (14 - 3) 2 2 1 1 a b y a ab x + = + − = Calculus with Substitution The second way to solve this problem is to recognize that this problem is a constrained optimization problem; a problem in which a function needs to be minimized while it’s being subjected to a constraint. The constrained minimization problem is converting into an unconstrained minimization problem. This is done using a substitution step. The constraint, Eq. (14 - 2), is substituted into the minimizing function, Eq. (14 - 1), to get (14 - 4) 2 2 ) ( b ax x J + + = The minimum of J is now found by taking the derivative of J with respect to x and setting it to zero. This yields MAE 461: DYNAMICS AND CONTROLS LINEAR ALGEBRAIC EQUATIONS (14 - 5) a b ax x ) ( 2 2 + + = which again leads to the answer given in Eq. (14 - 3). Calculus with No Substitution The third way of solving this problem is also done by converting the constrained minimization problem into an unconstrained minimization problem. However, this time, no substitution step will be needed to create the unconstrained minimization problem. This third method, which is the method that we’ll later employ to solve linear algebraic equations, proceeds by first writing the constraint as (14 - 6) = − − = b ax y f We then define the...
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achapt 14 - LINEAR ALGEBRAIC EQUATIONS 14 LINEAR ALGEBRAIC...

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